Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$

These are symmetric equations, so denoting $s=x+y,p=xy$ we have $$s^2-2p=2p^2$$ $$s(1+p)=4p^2$$ Now eliminate $s$: $$s^2=2p^2+2p=\left(\frac{4p^2}{1+p}\right)^2$$ $$2p(1+p)^3=16p^4$$ If $p=0$ then $s=0$ and obviously $x=y=0$. Otherwise, divide by $p$: $$(1+p)^3=(2p)^3$$ Assuming we are only solving in the reals: $$1+p=2p\qquad p=1,s=2$$ By Viète's formulas, $x$ and $y$ are the roots of $t^2-2t+1$, whereby we get $x=y=1$.

The first equation is equivalent to $(x+y)^2=2xy(1+xy)$.

Now multiplying the second equation by the previous relation yields: $$(x+y)^3=(2xy)^3$$ thus $x+y=2xy$ and then: $$(x+y)^2=2\times 2x^2y^2=2(x^2+y^2)\iff (x-y)^2=0$$

\begin{align*} (x+y)^2(1+xy)^2&=(4x^2y^2)^2\\ (x^2+y^2+2xy)(1+xy)^2&=16x^4y^4\\ (2x^2y^2+2xy)(1+xy)^2&=16x^4y^4\\ 2xy(1+xy)^3&=16x^4y^4 \end{align*} So, $xy=0$ or $1+xy=2xy$.

$xy=0$ or $1$.

If $xy=0$, $x=y=0$.

If $xy=1$, $x=y=1$.